Download Orbits - CSE-IITK

Lecture 4: Orbits
Rajat Mittal
?
IIT Kanpur
In the beginning of the course we asked a question. How many different necklaces can we form using 2
black beads and 10 white beads? In the question, the numbers 2 and 10 are arbitrarily chosen. To answer
this question in a meaningful way, we need to construct a strategy or theorem which will answer the above
question for any such numbers. But to understand the question better, lets ask a simpler question.
Exercise 1. How many different necklaces can be formed using 2 white and 2 black beads?
Lets look at the question in detail. The first guess for the question would be 4! = 24, the number of
ways we can permute the four beads. But all these permutations need not be different. What do we mean
by different necklaces? It might happen that two different permutations (σ1 and σ2 ) might be the same in
the sense that σ1 can be obtained from σ2 using rotation. Look at figure 1 for one such example.
Fig. 1. Two permutations giving the same necklace
Using some brute force now, we can come up with all possible different necklaces for 2 white and 2 black
beads (figure 1).
Fig. 2. Possible different necklaces with 2 white and 2 black beads
You can convince yourself that the question is much harder if we take bigger numbers. What should we
do? When are two necklaces equivalent?
Two necklaces are equivalent if we can obtain one by applying a symmetry to other necklace (like rotation
or reflection). We know from discussions in previous classes that the symmetries form the dihedral group,
D2n . The strategy would be to develop a general framework for groups to answer question about distinct
necklaces.
1
Group action
The first thing to notice in the necklace problem is that there are two different objects of interest. One is the
set of necklaces (set of all permutations of the necklaces) and other is the set of symmetries. A symmetry
can be applied to a necklace to obtain another necklace. Lets make this action abstract.
?
Thanks to the book from Dummit and Foote and the book from Norman Biggs.
Abstractly, given a group G and a set A, every element g of G acts on set A. That means for every
element g there is a function from A to A which is called its action on G. For the sake of brevity, we will
denote the function corresponding to the element g ∈ G with g itself. Hence the value of a ∈ A after action
of g will be called g(a).
Exercise 2. What is the group and what is the set for the necklace problem?
Note 1. It is NOT the set of distinct necklaces.
Lets look at the formal definition.
Definition 1. Given a group G and a set A, a group action from G to A assigns a function g : A → A for
every element g of group G. A valid group action satisfies the following properties.
– Identity takes any element a ∈ A to a itself, i.e., e(a) = a for every a ∈ A.
– For any two group elements g1 , g2 ∈ G, their functions are consistent with the group composition,
g1 (g2 (a)) = (g1 g2 )(a).
Using this definition and group structure of G, it can be shown that action of g is a permutation on the
elements of A.
This gives us another representation of group elements. For any group action on A of size m, we have a
permutation representation for any element g ∈ G in terms of a permutation on m elements. In the following
sections we will keep this representation in mind.
Note 2. Actually a slightly stronger theorem holds. It is called Cayley’s theorem and is given below. We will
not show the proof of this theorem.
Theorem 1. Cayley’s theorem: Every group of order n is isomorphic to some subgroup of Sn .
2
Orbits
Suppose we are given action of group G on a set A. Lets define a relation between the elements of A. If
∃ g : g(x) = y then we will say that x, y are related (x ∼ y). We can easily prove that this relation is
equivalence relation.
– Reflexive: Why?
– Symmetric: Suppose x ∼ y because g(x) = y. Then consider x = ex = (g −1 g)(x) = g −1 y, implying
y ∼ x.
– Transitive: Show it as an exercise.
Hence this equivalence relation will partition the set A into distinct equivalence classes. The equivalence
class corresponding to x ∈ A is the orbit (G(x)) of element x. In other words,
G(x) = {g(x) : g ∈ G}.
Now we will look at two counting questions,
1. What is the size of these orbits?
2. How many distinct orbits are there?
Why are we interested in these questions. Let us look at this concept from the example of necklaces. If a
necklace x can be obtained from another necklace y using a symmetry then they are related (in the necklace
case indistinguishable).
Exercise 3. Convince yourself that the number of distinct necklaces is the same as the number of distinct
orbits (equivalence classes) under the dihedral group D2n .
We will answer both the counting questions under the general group-theoretic framework. As a special
case, this will solve the necklace problem.
2
2.1
stabilizers
Remember that the orbit of x ∈ A under the action of G can be defined as,
G(x) = {g(x) : g ∈ G}.
If every g ∈ G took x to a different element, the size of the orbit would be |G|. But this is too much to
expect. If we consider any example, there will be lots of g ∈ G which will take x to a single element y. Lets
define this set as G(x,y),
G(x, y) = {g ∈ G : g(x) = y}.
Exercise 4. Does the set G(x, y) form a subgroup of G? Under what condition will it form a subgroup?
The answer to the previous exercise is when x = y. The set Gx := G(x, x) is called the stabilizer of x,
Gx = {g ∈ G : g(x) = x}.
Exercise 5. If you were not able to solve the previous exercise, prove that Gx is a subgroup of G.
Once we have the subgroup Gx , the natural question to ask is, what are the cosets? This is where we get
lucky. Suppose y is an element of the orbit G(x). So there exist an h ∈ G, s.t., h(x) = y. Then G(x, y) is
precisely the coset hGx .
Lemma 1. Given a y ∈ A, s.t., h(x) = y. The coset hGx is same as the set G(x, y).
Proof. ⇒: An element of hGx is of the form hg, g ∈ Gx . Then hg(x) = h(x) = y. So hGx ⊆ G(x, y).
⇐: Suppose g ∈ G(x, y), i.e., g(x) = y. Then show that,
Exercise 6. h−1 g ∈ Gx
But h−1 g ∈ Gx implies g ∈ hGx .
Exercise 7. Show the above implication. Be careful, It is not just the same as multiplying by h on both sides.
From the previous exercise G(x, y) ⊆ Gx .
Hence, for every element y in the orbit G(x), there is a coset. It is an easy exercise to convince yourself
that every coset will correspond to a single element in the orbit G(x). So the number of elements in the orbit
is equal to the number of cosets. But we know that the number of cosets can be calculated from Lagrange’s
theorem. Hence,
|G| = |Gx ||G(x)|.
Note 3. Gx is a subset of G, but G(x) ⊆ A. The equation works because we show a one to one relation
between G(x) (orbit) and the cosets.
2.2
Burnside’s lemma
We now know the size of the orbit. Given an element x with stabilizer Gx , the number of elements in its
|G|
orbit is |G
. Can this help us in counting the number of distinct orbits.
x|
1
Lets give every element on A a weight of |G(x)|
. The number of distinct orbits is now the sum of weights
of all elements of A.
Number of distinct orbits =
X
x∈A
3
1
1 X
=
|Gx |.
|G(x)|
|G|
x∈A
Lets concentrate on the summation. The total summation is equal to the number of pairs (g ∈ G, x ∈ A),
s.t., g(x) = x. Suppose we make a matrix with rows indexed by elements of G and columns indexed by
elements of A. The entry (g, x) is one if g(x) = x and 0 otherwise.
x1 x2 · · · x|A|
g1 0 1 · · · 1
g2 0 0 · · · 1
.. .. .. . .
.
. ..
. . .
g|G| 1 0 · · · 0
P
Then
x |G(x)| is the number of 1’s in the matrix above. Each term in the summation, |Gx | is the
number of 1’s in the column corresponding to x. We can count the number of 1’s in the matrix by taking
the sum row-wise too. Suppose, S(g) is set of elements of A fixed by g.
S(g) = {x : g(x) = x, x ∈ A}.
Using S(g) we get the orbit-counting (Burnside’s) lemma.
Lemma 2. Burnside’s lemma (Orbit-counting): Given a group action of G over A. The number of distinct
orbits can be written as,
Number of distinct orbits =
1 X
|S(g)|.
|G|
g∈G
Note 4. The summation is now over G instead of A.
One natural question you might ask is, How did this benefit us? Previously we were summing over all
possible x ∈ A and now we are summing up over all g ∈ G. The reason is, in general, the size G will be much
smaller than size of A.
Lets look at an example where Orbit-counting lemma will help us in answering the question about
necklaces. Try to solve this exercise yourself first and later you can look at the solution given below.
Exercise 8. How many necklaces can be formed with 2 black and 6 white beads?
Arrange the beads on the vertices of a regular 8-gon. Since the necklaces are obtained by fixing the
position of 2 black beads, there are 28 elements in A. The symmetry group is D16 with 16 elements.
For different elements of G we can calculate the number of elements fixed by it.
– Identity e: Fixes 28 elements.
– Out of 7 other rotations, only one of them fixes 4 elements. What is the angle of that rotation? Rest do
not fix anything.
– All reflections fix exactly 4 elements. It is easy to see by looking at the cycle structure of the permutation.
All beads of the same color should fall in the same cycle.
So by Orbit-counting lemma,
Number of distinct orbits =
1
(28 + 4 + 8 × 4) = 4.
16
Exercise 9. What are the four configurations? Can you characterize them?
For other examples of application of Burnside’s lemma, please look at section 21.4 of Norman Biggs book.
There is a nice example in Peter Cameron’s notes on Group Theory too (section 1.3).
4
3
Group representations (advanced)
We looked at the permutation representation of every group. There is a matrix representation of every group
too. The study of that representation is called the group representation theory. Group representation theory
is one of the main tools to understand the structure of a group. We will only give a very basic idea of this
field. Interested students can look at book Algebra by Artin.
Define GLn to be the group of invertible matrices of size n × n with complex entries. We can also think
of these matrices as linear operators over Cn .
A linear representation (matrix representation) is a homomorphism from group G to the group GLn (say
R : G → GLn ). That means, we map every element of group G to an invertible matrix, s.t., it obeys the
group composition,
R(gh) = R(g)R(h).
If there is a subspace of Cn which is fixed by every group element, then the representation is reducible.
In other words, for an irreducible representation, there is NO subspace which is fixed by every element of
group G.
Exercise 10. What does it mean that the subspace is fixed?
It can be shown that every representation can be broken down into irreducible representations. Another
quantity of interest is the character. The character of the representation is a function χ : G → C defined by
χ(g) = trace(R(g)).
The characters of irreducible representations are orthonormal to each other and satisfy various other
nice properties. Many theorems in group theory are derived by studying the characters of the group. Again,
interested students can find more information in the book Algebra by Artin.
4
Assignment
Exercise 11. Show that action of dihedral group on the set of necklaces is a group action.
Exercise 12. Show that G is isomorphic to a subgroup of S|A| . Remember that Sn is the group of permutations
of [n].
Hint: First show that action of g on A is a permutation and then use the consistency of group action
with the group composition.
Exercise 13. Suppose we want to find the number of necklaces with m black and n white beads. What is the
size of G and what is the size of A in terms of m, n?
Exercise 14. Prove that the average number of elements fixed by an element of group G under group action
is an integer.
Exercise 15. Prove that GLn is a group. What composition rule did you use?
Exercise 16. Biggs: Let G be a group of permutations of set A. If u, v are two elements in the same orbit of
G, show that |Gu | = |Gv |.
Exercise 17. Biggs: Let A denote the set of corners of the cube and let G denote the group of permutations
of A which correspond to rotation of the cube. Show that,
– G has just one orbit.
– For any corner x, |Gx | = 3.
– |G| = 24
Exercise 18. Biggs: Suppose you manufacture an identity card by punching two holes in an 3 × 3 grid. How
many distinct cards can you produce. Look at the figure given below.
Hint: The group to consider here is D8 .
5
Fig. 3. Different identity cards with circles showing the holes.
6